Hi,

this is again a question from me which did not get any answer at math.stackexchange (Link: http://math.stackexchange.com/questions/27366/)

This question is about how well one can choose a partition of unity.

So suppose $(M,g)$ is an open Riemannian manifold. I would like to have the following statement to be true:

"There exists an open cover $\{U_n\}$ of $M$ which is uniformly locally finite (i.e. each point of $p \in M$ lies in at most $q$ sets) and uniformly bounded (i.e. $\mathcal{U} := \bigcup U_n \times U_n$ is a controlled subset of $M \times M$, i.e. $\sup_{p \in \mathcal{U}} d(\pi_1(p), \pi_2(p)) < \infty$, where the $\pi_i$ are the projections on the first and second coordinate).

Furthermore, there exists an subordinate partition of unity $\{g_n\}$ with the properties that the functions $\{g_n^{1/2}\}$ are also smooth and for every $l \in \mathbb{N}$ the $i$-th derivatives $(g_n^{1/2})^{(i)}$ are uniformly bounded for all $i \le l$, i.e. $\parallel(g_n^{1/2})^{(i)}\parallel_\infty := \sup_{p \in M} |\nabla_{v_1, \ldots, v_i} g_n^{1/2}(p)| \le G_l$, where the $v_i$ are unit vectors at the point $p \in M$."

I think the most critical point is the uniform boundedness of the derivatives. Maybe someone knows a reference where it is proven that such a partition of unity always exists (or maybe just one, which has only the property of uniform boundedness of the derivatives)? Is this statement even true?

Thanks, Alex

edit: Though the metric $g$ on $M$ is fixed (in the application where I need this) I would be also happy about a solution of the form "There exists a metric $g$ on $M$, such that ... (the above holds)" or by giving some sufficient conditions on the metric, such that it is always possible.